Question Number 103168 by mohammad17 last updated on 13/Jul/20
$${The}\:{particular}\:{solution}\:{of}\:{differential}\:{equation}\: \\ $$$${of}\:\frac{{dy}}{{dx}}+\frac{{y}}{{x}}={k}\:{is}\:{y}=\frac{\mathrm{1}}{{x}}+\mathrm{2}{x}\:{thus}\:,{whats}\:{the}\:{value}\:{of}\:{k}\:? \\ $$
Answered by bemath last updated on 13/Jul/20
$${IF}\:{u}\left({x}\right)=\:{e}^{\int\:\frac{{dx}}{{x}}} \:=\:{e}^{\mathrm{ln}\left({x}\right)} \:=\:{x} \\ $$$${x}.\frac{{dy}}{{dx}}\:+\:{y}\:=\:{kx}\: \\ $$$$\frac{{d}}{{dx}}\:\left({xy}\right)\:=\:{kx}\: \\ $$$${xy}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{kx}^{\mathrm{2}} \:+\:{C}\: \\ $$$${y}\:=\:\frac{{kx}}{\mathrm{2}}\:+\:\frac{{C}}{{x}}\:,\:{then}\:\frac{{kx}}{\mathrm{2}}\:=\:\mathrm{2}{x} \\ $$$$\therefore\:{k}\:=\:\mathrm{4}\: \\ $$
Commented by mohammad17 last updated on 13/Jul/20
$${thank}\:{you}\:{sir} \\ $$